Conical train wheels I've been reading about how the conical shape of train wheels helps trains round turns without a differential. For those who are unfamiliar with the idea, the conical shape allows the wheels to shift and slide across the tracks, thus effectively varying their radii and allowing them to cover different distances while rotating at the same angular velocity. 
A cross-sectional view of the tracks and wheels generally looks something like:

But what about a configuration like the following?

I read in an online article that wheels in the second configuration may more easily slip and derail from the tracks (assuming there are no flanges to prevent them from doing so). But I can't convince myself using physics why that might be. 
Is one of these two configurations actually more reliable than the other?
 A: The contact with the rail creates a kinematic center of rotation where the reaction forces meet. The rail car will tend to rotate about this center as a result of side loads. 


*

*If the center is above the center of mass, the rail car acts like a hanging pendulum. A small deflection will cause a restoring torque opposing the swing.

*If the center is below the center of mass, the rail car acts like an inverted pendulum. A small deflection will cause a positive feedback amplifying the swing.

As a side effect the rail car will turn away from the turn instead of into the turn when the cone is the other way around.
A: Shift the upper configuration to the left a short distance at equilibrium. Result: the left wheel goes a little up, the right goes a little down, the train tilts clockwise, the center of mass is to the right of the centerline between the wheels, and therefore the center of mass provides a restorative force to push the train back to the right.
Shift the lower configuration to the left a short distance at equilibrium. The argument proceeds in reverse and the center of mass provides an anti-restorative force, pushing the train further to the left. Pain ensues.
You're trading $m \ddot x = - k x$ (harmonic oscillator) for $m \ddot x = k x$ (exponential diverger) and praying that the implicit drag forces keep the thing diverged only a small amount. That's a risky game, no doubt.
A: In both diagrams in the question, the left wheel has a smaller radius at the contact point than the right wheel. Because they're fixed to a common axle, in any given amount of time, the right wheel will travel a greater distance than the left, so the axle as a whole will rotate anti-clockwise (when viewed from above) about a vertical axis. As it does this, it will start to lie diagonally across the track, rather than perpendicular to the two rails.
Suppose we're using the wheel profile in the first diagram. As the axle gets farther from perpendicular, the right wheel moves ahead and drops down to have a lower-radius part in contact with the rail. This means that the axle is self-straightening since the more it steers, the more it tends to push itself back towards being perpendicular.
However, if we use the second wheel profile, as the axle gets farther from perpendicular, the right wheel climbs up the rail, causing the contact point to move to an area of higher radius. That means the right wheel goes even farther in one revolution, so it turns even more. That's completely unstable.
In the second diagram, the only way the axle can return to running straight is for the whole axle to rotate clockwise about a horizontal axis, with the wheels sliding perpendicularly across the rails. That would cause a huge amount of wear to both the wheels and the track, assuming the thing didn't just derail.
There's a Numberphile video demonstrating all of this with a few taped-together espresso cups.
A: To see why the first configuration is used rather than the second, perform the following experiment:
Hold a bowl in your hand and place a small ball inside. Move the bowl in circles at various speeds and observe the behavior of the ball.
Now turn the bowl over and balance the ball on top. Again, move the bowl around and observe the ball.
Which is more stable?
In both cases, there is a low center of mass, the motion of which is constrained by the geometry of the parts. In the first of each case - your first diagram of train wheels, or the upright bowl - the geometry pits the rotations caused by centrifugal forces against gravity; as the train wheels slide off-center, the center of mass is lifted up, and when gravity pulls it back down it centers itself again. In the second of each case, a lateral movement causes the center of mass to slide "downhill," and gravitational acceleration exacerbates the problem rather than correcting it.
A train with the first configuration of wheels keeps itself on the track naturally; the second configuration would require an expensive and challenging auto-ballast system to stay upright even when stationary.
A: Previous answers are great and explain the dynamics very well.  I'd like to point out that this can be explained just as easily in a static situation. 
Imagine the weight the shaft has to carry.  You don't even have to imagine a curve to note that the weight will automagically center (and lower the center of gravity of) the train in the first image. 
In the bottom image the flex of the shaft will cause the wheels to straighten out (thus raising the center of gravity). 
As previously mentioned, even if you could lower the center enough not to cause instability, the wear and tear on both wheels and tracks due to constant "climbing" would be reason enough not to proceed...
That and the typical "This is how we've always done it.  Don't think, just do." ;)
RE: dynamic situation: Now as the train rounds the curve and the centripedal (centrifugal?) force tries to topple the car, the conical wheels actually raise one side of the car while lowering the other thus moving the center of gravity closer to the pivot point. Also note that the shaft will flex more due to weight distribution on the inner wheel than on the outter. This will increase the angle of contact on the inner and decrease the outter counteracting the Newton's first law. Win, win, win.  
A: I can see a difference by thinking about the flanges...
Given that 
1) the flanges are on the inside of the wheels 
2) the right handside of the diagrams is the outer part of the track where the flange will press against the rail...
compare

and 

where the red lines indicate the plane of the flange.... in the upper case the flange neatly pushes against the edge of the rail and will not easily jump past it, but in the lower case the flange looks like it could slide up the outwardly angled rail and cause the train to derail.
Edit after interesting comment - if the flanges were on the outside of the wheels then the lower case would be potentially as good as the upper case with the flanges on the inside. I am not sure how easy it would be to make points etc. work with flanges on the outside...
...the main advantage I can see with inner flanges is that when building the railway lines a suitable length rod between the rails can check the gap between them that might be easier than having some large calipers with jaws to check the distance between the outer edges of the rails. There may be other challenges as suggested by JonCuster
So as noted above this answer assumes that the wheel flanges are inside the wheels and not outside.
